Can a symmetric positive definite matrix interpolation have linear trace and determinant?

Question

Definition. Suppose $A_0, A_1 \in S_+^N$ are real symmetric positive definite $N \times N$ matrices. An interpolation from $A_0$ to $A_1$ is a continuous / smooth function $A \colon [0, 1] \to S_+^N$ such that $A(0) = A_0$ and $A(1) = A_1$.

Example 1. The linear interpolation $A_{\text{lin}}(t) := (1 - t) A_0 + t A_1$ has linear trace, that is, $\text{tr}(A_{\text{lin}}(t)) = (1 - t) \text{tr}(A_0) + t \text{tr}(A_1)$, but not linear determinant: $\det(A_{\text{lin}}(t)) \ne (1 - t) \det(A_0) + \det(A_1)$.

Example 2. The logarithmic interpolation $A_{\log}(t) := \exp\left( (1 - t) \log(A_0) + t \log(A_1)\right)$ has linear determinant but not linear trace.

My question.

Does there exist an interpolation from $A_0$ to $A_1$ with linear trace and determinant?

My attempt. I tried to show that this is not possible for $N = 2$ (it clearly is possible for $N = 1$). Suppose $A_k = \begin{pmatrix} a_k & b_k \\ b_k & c_k \end{pmatrix}$ for $k \in \{ 0, 1 \}$ and $A(t) = \begin{pmatrix} a(t) & b(t) \\ b(t) & c(t) \end{pmatrix}$. Then we require that \begin{gather*} a(t) + c(t) \overset{!}{=} (1 - t) (a_0 + c_0) + t (a_1 + c_1) \\ a(t) c(t) - b(t)^2 \overset{!}{=} (1 - t) [a_0 c_0 - b_0^2] + t [a_1 c_1 - b_1^2]. \end{gather*}

We can't simply choose $a(t) = (1 - t) a_0 + t a_1$ and $c(t)$ similarly, because then we are in the case of Example 1 and thus don't have linear determinant.

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Since trace and determinant both only depend on the eigenvalues, there might be an argument using diagonalization.

Plugging the first into the second equation of @AndreasLenz yields $$ \lambda_1(t) \big((1 - t) (\lambda_1(0) + \lambda_2(0)) + t (\lambda_1(1) + \lambda_2(1)) - \lambda_1(t) \big) = (1 - t) \lambda_1(0) \lambda_2(0) + t \lambda_1(1) \lambda_2(1) $$ and thus \begin{align} 2 \lambda_1(t) & = (1 - t)(\lambda_1(0) + \lambda_2(0)) + t (\lambda_1(1) + \lambda_2(1)) \\ & \quad \pm \sqrt{((1 - t)(\lambda_1(0) + \lambda_2(0)) + t (\lambda_1(1) + \lambda_2(1)))^2 - 4 \lambda_1(0) \lambda_2(0) (1 - t) - 4 \lambda_1(1) \lambda_2(1) t}.\end{align}

Answer 1

This is not always possible. E.g. let $A_0=I_2$ and $A_1=2I_2$. If we want the trace and determinant of $A(t)$ to vary linearly with $t$, we need $$ \begin{aligned} \operatorname{tr} A(t)&=(1-t)(2)+t(4)=2(1+t),\\ \det A(t)&=(1-t)(1)+t(4)=1+3t.\\ \end{aligned} $$ The characteristic polynomial of $A(t)$ is therefore $p_t(x)=x^2-2(1+t)x+(1+3t)$. However, when $t\in(0,1)$, the discriminant of $p_t$, $\big(2(1 + t)\big)^2 - 4 (1 + 3 t) = 4 t (t - 1)$ is negative. Hence $A(t)$ cannot even be real symmetric on this open interval, not to say positive definite.

Answer 2

Some definitions first. Define $p_N:[0,1]\to[0,1],\,p_N^\max\in\mathbb R$ and $\tau,\delta:[0,1]\to\mathbb R_{>0}$ by $$ \begin{aligned} p_N(x)&=\dfrac{x(1-x)^{N-1}}{(N-1)^{N-1}}\quad\text{(with $p_1(x)=x$ by convention),}\\ p_N^\max&=\max\limits_{x\in[0,1]}p(x)=p\left(\dfrac1N\right)=\dfrac{1}{N^N},\\ \tau(t)&=(1-t)\operatorname{tr} A_0+t\operatorname{tr} A_1,\\ \delta(t)&=(1-t)\det A_0+t\det A_1.\\ \end{aligned} $$ Here is a partial answer to your question:

$$ \bbox[10px,border:2px solid red] { \text{A continuous solution $A$ exists if and only if $\dfrac{\delta}{\tau^N}\le p_N^\max=\frac{1}{N^N}$ on $[0,1].\qquad(\dagger)\ $} } $$ Its necessity is an easy consequence of the AM-GM inequality for the eigenvalues of $A(t)$. We now prove its sufficiency.

Let $A_i=e^{K_i}S_ie^{-K_i}$ be an orthogonal diagonalisation, where $K_i$ is skew-symmetric and $S_i$ is a positive diagonal matrix. If we can find a continuous positive diagonal matrix function $S(t)$ such that $S(0)=S_0,\,S(1)=S_1,\,\operatorname{tr}S(t)=\tau(t)$ and $\det S(t)=\delta(t)$ where $\tau$ and $\delta$ are the linear interpolation functions defined at the beginning of this answer, a continuous solution $A(t)$ is obtained: $$ \begin{cases} K(t)=(1-t)K_0+tK_1\\ A(t)=e^{K(t)}S(t)e^{-K(t)} \end{cases}. $$ Let $X=\dfrac{1}{\tau}S$. Whenever $X(t)$ exists, we have $\operatorname{tr}X(t)=1$ and $0<f_N(t):=\det X(t)=\dfrac{\delta(t)}{\tau(t)^N}\le p_N^\max$ for all $t$, where the latter inequality is due to $(\dagger)$. Let $x_i(t)$ be the $i$-th diagonal element of $X(t)$. The existential problem of $S$ thus reduces to the following abstract problem:

• Given any continuous function $f_N:[0,1]\to\mathbb R_{>0}$ such that $f_N\le p_N^\max$ and given any $2N$ real numbers $x_1(0),x_2(0),\ldots,x_N(0),x_1(1),x_2(1),\ldots,x_N(1)$ such that for $T\in\{0,1\}$, we have $$ 0<x_i(T)\leq 1 \text{ for all $i$},\quad \sum_{i=1}^Nx_i(T)=1\quad \text{and}\quad\prod_{i=1}^Nx_i(T)=f_N(T),\tag{$\ast$} $$ is it possible to extend each $x_i$ to a continuous functions $x_i:[0,1]\to(0,1]$ such that condition $(\ast)$ is satisfied when $T\in\{0,1\}$ is replaced by any $t\in[0,1]$?

We shall prove the feasibility of such extension by mathematical induction on $N$. The base case $N=1$ is solved by the constant function $x_1=1$. In the inductive case, suppose $N\ge2$. When $T=0,1$, by the AM-GM inequality, we have $$ f_N(T) =\prod_ix_i(T) \le x_N(T)\left(\frac{\sum_{i<N}x_i(T)}{N-1}\right)^{N-1} =x_N(T)\left(\frac{1-x_N(T)}{N-1}\right)^{N-1} =p_N(x_N(T)). $$ Since $f_N\le p_N^\max$, there exists some continuous function $h$ on $[0,1]$ such that $f_N\le h\le p_N^\max,\ h(T)=p_N(x_N(T))$ for $T\in\{0,1\}$ and $h(\frac1N)=p_N^\max$.

Note that $p_N$ is a continuous function that is strictly increasing on $\mathcal I=[0,\frac1N]$ and strictly decreasing on $\mathcal J=[\frac1N,1]$. So, for $T=0,1$, if we define $$ q_T=\begin{cases} (p_N|_{\mathcal I})^{-1}&\text{if }h(x_N(T))\in[0,\frac1N],\\ (p_N|_{\mathcal J})^{-1}&\text{if }h(x_N(T))\in(\frac1N,1],\\ \end{cases} $$ then $q_T\circ h:[0,1]\to[0,1]$ is a continuous function such that $(q_T\circ h)(T)=x_N(T)$. Therefore we may extend $x_N$ to a continuous function on $[0,1]$ by defining $$ x_N(t)=\begin{cases} q_0\circ h&\text{if }t\in[0,\frac1N],\\ q_1\circ h&\text{if }t\in(\frac1N,1].\\ \end{cases} $$ It follows that $p_N(x_N(t))=h(t)\ge f_N(t)>0$. Since $p_N(0)=p_N(1)=0$, we must have $0<x_N(t)<1$ on the unit interval. We may therefore define a continuous function $f_{N-1}:[0,1]\to\mathbb R_{>0}$ by $$ f_{N-1}(t)=\frac{f_N(t)}{x_N(t)(1-x_N(t))^{N-1}}. $$ Then $$ f_{N-1}(t) \le\frac{h(t)}{x_N(t)(1-x_N(t))^{N-1}} =\frac{p(x_N(t))}{x_N(t)(1-x_N(t))^{N-1}} =\frac{1}{(N-1)^{N-1}} =p_{N-1}^\max. $$ and hence $0<f_{N-1}\le p_{N-1}^\max$. Now for $T=0,1$ define $y_i(T)=\frac{x_i(T)}{1-x_N(T)}$ for $i=1,2,\ldots,N-1$. One may readily verify that $0<y_i(T)\leq 1,\ \sum_{i=1}^{N-1}y_i(T)=1$ and $\prod_{i=1}^{N-1}y_i(T)=f_{N-1}(T)$. By induction hypothesis, we may extend the $y_i$s to to continuous functions on $[0,1]$ such that the three properties of the $y_i$s conditions are true when $T$ is replaced by any $t\in[0,1]$. Now define $x_1,x_2,\ldots,x_{N-1}$ by $x_i=y_i(1-x_N)$ and we are done.